JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Stopping of Vehicle by Retarding Force

If a vehicle moves with some initial velocity and due to some retarding force it stops after covering some distance after some time.

(1) Stopping distance :  Let m = Mass of vehicle, 

\[\upsilon \] = Velocity,   P = Momentum,  E = Kinetic energy

F = Stopping force,     x = Stopping distance,

t = Stopping time

Then, in this process stopping force does work on the vehicle and destroy the motion.

By the work- energy theorem

\[W=\Delta K=\frac{1}{2}m{{v}^{2}}\]  

\[\Rightarrow \] Stopping force (F) x Distance (x) = Kinetic energy (E)

\[\Rightarrow \] Stopping distance (x) \[=\frac{\text{Kinetic}\,\text{energy}\,\text{(}E\text{)}}{\text{Stopping}\,\text{force}\,\text{(}F\text{)}}\] 

\[\Rightarrow \] \[x=\frac{m{{v}^{2}}}{2F}\]                                          ...(i)

(2) Stopping time : By the impulse-momentum theorem

\[F\times \Delta t=\Delta P\Rightarrow F\times t=P\]

\[\therefore \] \[t=\frac{P}{F}\]

or \[t=\frac{mv}{F}\]                                                ...(ii)

(3) Comparison of stopping distance and time for two vehicles : Two vehicles of masses \[{{m}_{1}}\] and \[{{m}_{2}}\] are moving with velocities \[{{\upsilon }_{1}}\] and \[{{\upsilon }_{2}}\] respectively. When they are stopped by the same retarding force (F).

The ratio of their stopping distances \[\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{m}_{1}}v_{1}^{2}}{{{m}_{2}}v_{2}^{2}}\] and the ratio of their stopping time  \[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{m}_{1}}{{v}_{1}}}{{{m}_{2}}{{v}_{2}}}\]           

(i) If vehicles possess same velocities

\[{{\upsilon }_{1}}={{\upsilon }_{2}}\]

\[\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{m}_{1}}}{{{m}_{2}}}\]  ; \[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{m}_{1}}}{{{m}_{2}}}\]         

(ii) If vehicle possess same kinetic momentum         

\[{{P}_{1}}={{P}_{2}}\]

\[\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}=\left( \frac{P_{1}^{2}}{2{{m}_{1}}} \right)\,\left( \frac{2{{m}_{2}}}{P_{2}^{2}} \right)\,=\frac{{{m}_{2}}}{{{m}_{1}}}\]           

\[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{P}_{1}}}{{{P}_{2}}}=1\]           

(iii) If vehicle possess same kinetic energy           

\[\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}=1\]           

\[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{\sqrt{2{{m}_{1}}{{E}_{1}}}}{\sqrt{2{{m}_{2}}{{E}_{2}}}}=\sqrt{\frac{{{m}_{1}}}{{{m}_{2}}}}\]

Note :

If vehicle is stopped by friction then

Stopping distance \[x=\frac{\frac{1}{2}m{{v}^{2}}}{F}\]\[=\frac{\frac{1}{2}m{{v}^{2}}}{ma}\]\[=\frac{{{v}^{2}}}{2\mu g}\]      

\[[\text{As}\,\,a=\mu g]\]

Stopping time  \[t=\frac{mv}{F}\]\[=\frac{mv}{m\mu \,g}\]\[=\frac{v}{\mu \,g}\]